On Sep 2, 7:09 am, "Tim Golden BandTech.com" <tttppp...@yahoo.com> wrote: > > As I go over it again I will stand by the 2 pi claim. If you have a > falsification then please present it. I don't necessarily believe all > of mainstream mathematics so I find it acceptable that you do not > believe mine. Simply stated: the sum of the angles between three > coplanar vectors will always be 2 pi, in any dimension. The converse > holds: If the sum of the angles of three vectors is 2 pi, then the > vectors are coplanar. This is not a proof, but is an observation. I'm > open to falsification, but doubt that you will find one. > > - Tim
I don't really understand what you mean by "sum of the angles between three coplanar vectors will always be 2 pi." I'm assuming you meant dimension greater than or equal to 2 when you said "in any dimension." This statement is trivially false if you just meant "three vectors that that lie in a 2 dimensional subspace."(Take any vector X. Then, take 2X and 3X.) Even if you meant "three vectors that span a 2- dimensional subspace," it is trivially false. (Take two linearly independent vectors, X,Y. Choose X+Y as the third vector. Angle between X&Y, X&(X+Y), and Y&(X+Y) will not add up to 2Pi.)
So obviously, you mean something else. Exactly what do YOU mean by three co-planar vectors?
The converse of the statement is also trivially false. Look at a tetrahedron constructed from 4 equilateral triangles. Imagine that one of the vertices is the origin, and imagine that the three edges are vectors in R^3. What are the angles between the vectors? What is the sum of the angles of these vectors? Are these vectors "coplanar"?
I agree with Gerry that perhaps it would be wise for you to learn a bit of conventional mathematics before rejecting it. If you know some algebra, you'd know that your polysign numbers is just a example of a quotient ring.